Why would voltage decrease over a resistor?

[mpatryluk]

What happens to a circuit over a resistor that causes measured voltage to drop?
If voltage is related to a charge imbalance creating electric potential between two points, then what does electrons moving through a resistor have to do with a reduction of that charge imbalance?

Series circuit:

Diagram 1

Negative (10 volts)--> ----------------A1-----1 ohm resistor-----B1-------------Positive terminal

Diagram 2

Negative (10 volts)--> ----------------A2-----No resistor---------B2-------------Positive terminal




Since I is fixed for the entire circuit, it's the same at points A and B. A1 and B1 would have a lower I value than A2 and B2.

The only difference i can think of is that electrons give off energy in the resistor, so their energy levels would be lower at point B than point A. Does that mean that voltage measured is in relation to the energy of the electrons at a given point? How does that work? Is it supposed to represent how much of the electron's potential has been "used up" since the start of the circuit?

In diagram 2, there's no resistor, except for the marginal resistance of the wire, which we could consider to be much less than 1 ohm. But if voltage along a circuit is related to electron energy, and all series circuits go from full to 0 voltage over the course of the circuit, then the energy in the electrons in diagram 2 must reach the level of the electrons in diagram 1 by the end.

Diagram 2 has a larger current. Does this mean that the electrons lose energy more quickly along the course of the non resistive wire than they would over the same wire in a circuit where there was also a resistor? (i.e. diagram 1) So then increased current means that current will suffer a voltage drop more quickly over a fixed resistance than a slower current?

 

[phinds]

Since I is fixed for the entire circuit, it's the same at points A and B. A1 and B1 would have a lower I value than A2 and B2.

This is a self-contradictory statement. The only way I can be fixed over the entire circuit is if it is provided by an ideal current source, and if it IS, there is no difference in current between A1/B1 and A2/B2.

 

[rcgldr]

To avoid the issue of idealized zero resistance conductors, change this to:

Series circuit:

Diagram 1

Negative (10 volts)-->A1--------------------1.000 ohm resistor--------------B1>--Positive terminal

Diagram 2

Negative (10 volts)-->A1--------------------0.001 ohm resistor--------------B1>--Positive terminal

---

Assuming voltage is fixed, then there is 1000 times as much current flowing in diagram 2 as there is in diagram 1, but the voltage at A1 and B1 would be the same in both diagrams.

 

[jbriggs444]

What happens to a circuit over a resistor that causes measured voltage to drop?

There is an analogy that may help your intuition. Electric current is like the flow of an incompressible fluid under pressure. Like water in pipes.

Pressure is the analog to voltage. The pressure difference between two points in the pipe system is what causes the water to flow.

Flow rate (e.g. liters per second) is the analog to current. Since the water is incompressible then the flow at any point on a closed loop must be the same as at every other point.

A section of pipe that resists flow is the analog to a resistor. An ideal pipe section will pass a flow rate that is proportional to the applied pressure just like an ideal resistor will pass a current that is proportional to the applied voltage.

You can extend the analogy to capacitors and inductors. An inductor is like a mass of moving water in the pipes; it resists changes in flow rate. Make a coil of pipes to hold a lot of moving water and you've built an inductor. A capacitor is like a diaphragm in the pipe. It allows some flow to pass but builds up more and more pressure as a result until no more water can pass.

So, to answer your question by way of the analogy, the voltage drop across a resistor is like the pressure drop across a section of pipe.

 

[Delta Kilo]

First of all, your second diagram is a short circuit. Something is going to melt. Don't do it at home:)
Consider instead 2 resistors in series (voltage divider): [+3V]--a--1Ω--b --2Ω--c--[0V]

You are correct in saying that electric field is created by charge imbalance distribution. But it is only half of the story. Charge distribution is not a given. Each individual charge is affected by the overall electric field and it will move if it can. The final shape of the charge distribution is then determined by the equilibrium conditions. The actual shape can be quite complicated (it depends on circuit geometry), thankfully we don't need to worry about it. Just the fact that the equilibrium is reached is sufficient. To maintain the equilibrium, the number of electrons in must be = number of electrons out for any part of the circuirt (Kirchhoff's law). As a consequence, when connected in series, the current through each element is the same.

In conductors it takes some efforts to pull electrons throught the crystal lattice and the resulting flow of electrons is proportional to the applied force.(this is grossly oversimplified, see Ohm's law - miroscopic origins )

Let's say initially the voltage is the same across both resistors. Then the initial flow of electrons through Rab is going to be 2 times more than through Rbc, causing lack of electrons at point b (positive net charge). This will raise the potential of point b, thereby reducing the voltage across Rab and increasing it across Rbc until the equilibrium is reached and the flow through both resistors is the same. The voltages and the current can then be found from Ohm's law Vab/1Ω=Vbc/2Ω=Ia=Ib=Ic. Since Vab+Vbc=Vac=3V, Vab=1V, Vbc=2V, Ia=Ib=Ic=1A

 

[phinds]

To avoid the issue of idealized zero resistance conductors, change this to:

Series circuit:

Diagram 1

Negative (10 volts)-->A1--------------------1.000 ohm resistor--------------B1>--Positive terminal

Diagram 2

Negative (10 volts)-->A1--------------------0.001 ohm resistor--------------B1>--Positive terminal

---

Assuming voltage is fixed, then there is 1000 times as much current flowing in diagram 2 as there is in diagram 1, but the voltage at A1 and B1 would be the same in both diagrams.

Yes, that's correct. What's your point?

 

[Naty1]

What happens to a circuit over a resistor that causes measured voltage to drop?

Resistance in electrical circuits is like friction in mechanical...they both dissipate energy in the form of heat. A conductor has large numbers easily displaced electrons, a resistor not so much, and an insulator has each electron is tightly bound to a single molecule..so a lot of energy [POWER = IE] is required for current to flow.

An easy way to visualize this is that when a resistor gets hot, there is a lot of motion among the electrons, thermal motion which we measure as heat.

If voltage is related to a charge imbalance creating electric potential between two points, then what does electrons moving through a resistor have to do with a reduction of that charge imbalance?,...

Nothing. Current is the same entering as departing any individual resistor. There is no 'charge imbalance' whatever that means. q = it. In a series, circuit the same charge flows uniformly. electrons do not mysteriously appear or disappear along the series circuit, but their energy does vary. It is the electric potential that varies, which we describe as the potential energy of the charge. An electron at a high potential is analogous to a high stone: both can do more work than a low one as it drops.


http://en.wikipedia.org/wiki/Electric_potential

The electric potential at a point is equal to the electric potential energy (measured in joules) of any charged particle at that location divided by the charge (measured in coulombs) of the particle. Since the charge of the test particle has been divided out, the electric potential is a "property" related only to the electric field itself and not the test particle.

 

[sophiecentaur]

I prefer to put it all down to the same Conservation Laws that apply everywhere else. I also think that introducing Fields into circuit discussion serves only to confuse. Potential is just as fundamental as Field (I have made this point many times) and takes the 'geometry' out of the problem so why not just discuss circuit matters in terms of Potential? Anyone who has a problem with that approach would be better employed understanding more about what Potential means than going in ever decreasing circles trying to explain the (literal) ups and downs that you need to consider when the physical layout of a circuit has to be included in the reasoning. (You can't have Volts per metre without considering the metres, remember.

 

[Naty1]

If you like the water analogy, then your question

If voltage is related to a charge imbalance creating electric potential between two points, then what does electrons moving through a resistor have to do with a reduction of that charge imbalance?,...

can be also addressed analogously to a falling stream of water from a high to a low point: The water, molecules if you like, have higher potential when they start, lower potential when they get closer to the bottom of their fall. The number of molecules, the volume of water remains the same, but potential energy of each diminishes.

 

[Delta Kilo]

I prefer to put it all down to the same Conservation Laws that apply everywhere else. I also think that introducing Fields into circuit discussion serves only to confuse. Potential is just as fundamental as Field (I have made this point many times) and takes the 'geometry' out of the problem so why not just discuss circuit matters in terms of Potential?

Well, the problem with conservation laws is they don't explain what actually happens in between. Like, how does the voltage from the battery over there cause electrons to flow through the light bulb over here? Does the voltage travel through the wires - no it does not, not really, although it sure looks like it does.

Keep in mind, the person was asking what the voltage IS. The usual definition of voltage is given in terms of electrostatics, but then it is applied to circuit theory, with no explanation. One is left with $V(r) =\frac{1}{4\pi ε_0}\frac{Q}{r}$ on one hand and $V=IR$ on the other and not a clue how the two are related.

I mean, one has to get the picture right at least once, to see how it all fits together. Once it's done, one can forget all about charge distributions and imbalances and just use conservation laws as you suggest.

 

[Naty1]

What happens to a circuit over a resistor that causes measured voltage to drop?

I wonder why this question seems so often asked and why so few seem to ask

"How did the electrons gain potential energy at the source...in a generator or a battery."

maybe it's because the resistor seems so 'passive', so 'simple'...

 

[sophiecentaur]

I wonder why this question seems so often asked and why so few seem to ask

"How did the electrons gain potential energy at the source...in a generator or a battery."

maybe it's because the resistor seems so 'passive', so 'simple'...

My theory about this is that people ask questions that are just on the fringe of their comprehension and in terms they already understand. Considering a battery is just more demanding than thinking about the 'downhill trip' through a resistor.
It's the result of expecting easy concrete answers for something as abstract as Electricity. Science has been sold short by teachers for a long time.

 

[sophiecentaur]

Well, the problem with conservation laws is they don't explain what actually happens in between. Like, how does the voltage from the battery over there cause electrons to flow through the light bulb over here? Does the voltage travel through the wires - no it does not, not really, although it sure looks like it does.

Keep in mind, the person was asking what the voltage IS. The usual definition of voltage is given in terms of electrostatics, but then it is applied to circuit theory, with no explanation. One is left with $V(r) =\frac{1}{4\pi ε_0}\frac{Q}{r}$ on one hand and $V=IR$ on the other and not a clue how the two are related.

I mean, one has to get the picture right at least once, to see how it all fits together. Once it's done, one can forget all about charge distributions and imbalances and just use conservation laws as you suggest.

Some comments on this.
Firstly, it is expecting far too much to know "what actually happens" or what a voltage "really is". There is a definition of Voltage that is the amount of energy (Joules) available per unit of charge (Coulomb). That is the fundamental way that the Volt is defined and used. It works equally well in circuits as in space.
I don't know where you got the idea that there is no 'explanation' relating to circuit theory. What books do you read? Your electrostatic potential formula refers to an isolated point charge and is way down the line from any serious definition.

 

[Naty1]

delta...

One is left with V(r)=14πε0Qr on one hand and V=IR on the other and not a clue how the two are related.

I mean, one has to get the picture right at least once, to see how it all fits together.

how do you see those two concepts fitting together??

 

[rcgldr]

Like, how does the voltage from the battery over there cause electrons to flow through the light bulb

For a battery or other electrical power source, there's a difference in the amount of charge on the terminals, and this differential in charge results in electron flow and voltage is related to this differential in charge. An ideal power source will maintain this differential of charge and voltage despite the amount of current flowing between the terminals, but for a battery, the voltage and the charge differential will be somewhat reduced if the amount of current being consumed is increased.

Similarly, in the case of a resistor, there is a differential in charge and voltage between the "terminals" of the resistor, but the resistor is consuming energy, converting it to heat. In a simple circuit with a battery and resistor, the voltage increases as current flows through the battery due to internal chemical reactions, and the voltage decreases as current flows through the resistor due to conversion of energy into heat.

 

[Naty1]

there's a difference in the amount of charge on the terminals

What does that mean??

edit: You mean there are excess electrons at one terminal?

 

[sophiecentaur]

For a battery or other electrical power source, there's a difference in the amount of charge on the terminals, and this differential in charge results in electron flow and voltage is related to this differential in charge. An ideal power source will maintain this differential of charge and voltage despite the amount of current flowing between the terminals, but for a battery, the voltage and the charge differential will be somewhat reduced if the amount of current being consumed is increased.

Similarly, in the case of a resistor, there is a differential in charge and voltage between the "terminals" of the resistor, but the resistor is consuming energy, converting it to heat. In a simple circuit with a battery and resistor, the voltage increases as current flows through the battery due to internal chemical reactions, and the voltage decreases as current flows through the resistor due to conversion of energy into heat.

That is a very misleading contribution. It's like saying it 's the mass of the locomotive that makes the train move. It's the Potential Difference that causes the charges to flow. Before you launch into an 'explanation' like that, you could at least look at a book or even Wiki. How can anyone hope to learn anything from PF when that nonsense is posted, posing as authority!

 

[Naty1]

That is a very misleading contribution.

exactly!

 

[mikeph]

That is a very misleading contribution. It's like saying it 's the mass of the locomotive that makes the train move. It's the Potential Difference that causes the charges to flow. Before you launch into an 'explanation' like that, you could at least look at a book or even Wiki. How can anyone hope to learn anything from PF when that nonsense is posted, posing as authority!

Beg your pardon but he's right. Where do you think an electric potential comes from if not a non-uniform charge distribution? It's Gauss' law.

 

[sophiecentaur]

Beg your pardon but he's right. Where do you think an electric potential comes from if not a non-uniform charge distribution? It's Gauss' law.

I see what you say BUT: The problem is that there's no useful quantitative relationship when you look at it that way. It's a matter of cause and effect. You could connect a 1pF capacitor, charged to 1V, across a 1k resistor. There would be 1pC of charge and the initial current would be 1mA. Connect it across a 1F capacitor and the initial current would still be 1mA but the charge would be 1C. So trying to relate current to charge difference is a non-starter. Gauss' law says nothing about Current, btw.

 

[mikeph]

That's why you use it in conjunction with Ohm's law and the continuity equation.If you think trying to relate charge difference to current is a non-starter then I imagine you will have some difficulty telling me anything about the RC circuit except for at t = 0.

 

[sophiecentaur]

That's why you use it in conjunction with Ohm's law and the continuity equation.


If you think trying to relate charge difference to current is a non-starter then I imagine you will have some difficulty telling me anything about the RC circuit except for at t = 0.

I do not know what you mean by that.

What I wrote, applies to any value of t. Just think about it rather than arguing. The charge (together with with Capacitance) determine the PD, which will determine the Current. Not the charge by itself. It is the PD which determines the current. Really. This is so fundamental. Why don't you just read a textbook and see what's going on here, instead of using home brewed ideas.
Btw, Ohm's Law is not a universal law of nature and only applies 'when it applies' i.e to metals at constant temperature. It is not the 'definition' of resistance, which is V/I. In many components, V/I is not a constant.
What do you mean by the "continuity equation"? I don't know of it by that name.

 

[Delta Kilo]

Firstly, it is expecting far too much to know "what actually happens" or what a voltage "really is".

I repsectfully disagree. There is a (grossly oversimplified) model of conduction where positively charged
little balls stay put and negatively charged little balls bounce around. This picture, along with Coulomb's law and a few simple assumptions about steady state, is sufficient to describe he behaviour of resistive circuits. Add to this the assumption of some non-electrical forces acting on charges inside the battery and the circuit is complete.

Starting with Coulomb's law one can introduce the notion of test charge and electric field. Since individual fields add up, one can then compute the total field as a sum over all charges or integral over the volume. By noting conservative nature of the force, potential can be introduced.
Applying steady state assumption, one can arrive at zero potential difference in perfect conductors. Balance of electic and non-electric forces leads to fixed voltage at battery terminals. Then, using microscopic derivation of the Ohm's law, one can work out current through a resistor and eventually get all the usual circuit laws.

Your electrostatic potential formula refers to an isolated point charge and is way down the line from any serious definition.

It's not mine :)
Anyway, consider a 1mm 1Ω resistor with a voltage drop of 1V. Where did the 1V potential and 1KV/m electric field come from? Well, it didn't come all the way from the battery for sure. Instead it came from the distribution of charges in and immediately around the resistor according to the above formula. How that particular charge distribution came about that's another question.

 

[sophiecentaur]

I repsectfully disagree. There is a (grossly oversimplified) model of conduction where positively charged
little balls stay put and negatively charged little balls bounce around. This picture, along with Coulomb's law and a few simple assumptions about steady state, is sufficient to describe he behaviour of resistive circuits. Add to this the assumption of some non-electrical forces acting on charges inside the battery and the circuit is complete.

Starting with Coulomb's law one can introduce the notion of test charge and electric field. Since individual fields add up, one can then compute the total field as a sum over all charges or integral over the volume. By noting conservative nature of the force, potential can be introduced.
Applying steady state assumption, one can arrive at zero potential difference in perfect conductors. Balance of electic and non-electric forces leads to fixed voltage at battery terminals. Then, using microscopic derivation of the Ohm's law, one can work out current through a resistor and eventually get all the usual circuit laws.

It's not mine :)
Anyway, consider a 1mm 1Ω resistor with a voltage drop of 1V. Where did the 1V potential and 1KV/m electric field come from? Well, it didn't come all the way from the battery for sure. Instead it came from the distribution of charges in and immediately around the resistor according to the above formula. How that particular charge distribution came about that's another question.

What you are saying, here, is that you can make people feel better about electricity by giving them a (as you yourself describe it) poor model. The only good thing about this model is that it uses terms they are familiar with. What is the point of dumbing down like that, if the model doesn't actually provide anything other than an illusion of understanding? To find answers and to predict things, it's necessary to use the Maths, which you can rely on and to be aware of the variables that are relevant to any particular problem. Electricity is not 'concrete' and you can't treat it as if it is.

I am not challenging Coulomb's Law and I only called it "your law" because you are trying to use it to prove a point. I say that it is not relevant to this argument.
I understand that there is a simplified model (a bit more sophisticated than the little balls model you were quoting) of what electrons do in a metal and that it will involve the local fields between electrons and nuclei and that there will be a 'surplus' of electrons at the negative terminal of a load, which is continually replenished by the power supply. You will be aware that the amount of this surplus needs only to be very 'small' for the flow to occur. That applies to the situation at every point round the circuit because Kirchoff 1 can be said to apply everywhere. We are dealing with the analogy to an incompressible fluid and any local 'storage' of charge is Capacitance.
As I stated before, the instantaneous discharge current from a capacitor (which holds charges on each plate) through a resistor is totally independent of the capacity (hence the charge). It just depends upon the instantaneous PD value.

I think you are confusing a perfectly correct formula which states the relationship between variables with a causal relationship that will actually deliver answers about the current in a circuit.
At a more basic level, I would say that it is Energy that makes things happen and Charge is not energy.

You say it is possible to derive the circuit laws in 'your way'. Can you show where this is done in a reference (or even give a page of detailed derivation)? It sounds like a lot of trouble to produce what would, in effect, be a circular argument and you would have, at some point, to introduce Energy - which is there from the beginning when you use Potential. Are al the textbooks wrong when they do it the conventional way?

 

[voko]

At a more basic level, I would say that it is Energy that makes things happen and Charge is not energy.

This is the same debate we frequently have here: "force vs energy".

The truth is, they are complimentary views. You can do physics using energy, or you can do it using force. Sometimes one is easier than another, but they are equivalent.

 

[sophiecentaur]

This is the same debate we frequently have here: "force vs energy".

The truth is, they are complimentary views. You can do physics using energy, or you can do it using force. Sometimes one is easier than another, but they are equivalent.

Energy and Force are related (complementary, if you like) so you can use either but Charge is a totally different entity. Charge would only be there because it put there by the application of a potential / field. The 'Charge' description doesn't yield consistent answers. There is no one-to-one relationship between the charge 'on either side' of a resistor and the current through it. Where exactly would this charge be located, in any case? Back on the plates of a capacitor, somewhere in the windings of a transformer, in the chemicals of a battery? It's a nonsense explanation and no one has given a proper reference to justify it.

 

[rcgldr]

Wiki articles:

In the case of a battery, charge separation that gives rise to a voltage difference is accomplished by chemical reactions at the electrodes

http://en.wikipedia.org/wiki/Electromotive_force

http://en.wikipedia.org/wiki/Battery_(electricity)

http://en.wikipedia.org/wiki/Differential_capacitance

 

[sophiecentaur]

Sorry but I don't see where that takes us.

 

[voko]

Energy and Force are related (complementary, if you like) so you can use either but Charge is a totally different entity.

I do not see it that way.

Charge is directly related for force. Pretty much like mass is directly related to force (of gravity).

A distributed mass creates a force field; equivalently it creates a potential field.

So does the charge.

You are right, however, when you point out that there may be something else that creates an electric force field or potential.

 

[Delta Kilo]

Where exactly would this charge be located, in any case? Back on the plates of a capacitor, somewhere in the windings of a transformer, in the chemicals of a battery? It's a nonsense explanation and no one has given a proper reference to justify it.

Here we go, 2nd hit on google, with pictures:
http://www.astrophysik.uni-kiel.de/~hhaertel/PUB/voltage_IRL.pdf

CONSTANT ELECTRIC CURRENT AND THE DISTRIBUTION OF THE SURFACE CHARGES¹
Hermann Härtel, Guest scientist at Institute for Theoretical Physics and Astrophysics, University Kiel
1.This is the revised part of the full paper published under the title "A Qualitative ApproachTo Electricity"at:Institute for Research on Learning, Report No. IRL87-000 1987"

ABSTRACT

Surface charges are present, whenever a wire carries an electric current and are necessary for a causal explanation of how an electric current actually works.

Sorry, I don't think I can find an article in peer-reviewed journal because frankly, it has all been sorted out a century ago, now it's kid's textbook level stuff.

And again, I'm not suggesting any kind of "alternative approach to electricity", oh no, certainly not. I'm just saying that there is a qualitative picture of the process, based on a simple model, to complement quantitative methods using conservation laws.

It's a bit like colliding billiard balls. Ordinarily one solves the problem by applying energy and momentum conservation. The forces are not easily quantifiable and do not appear anywhere in the equations used to solve the problem. Nevertheless, if the question was "why did the ball change direction" the answer would be "because of the force acting on it" in line with the Newton's 1st law.

And btw, this picture, while simplified, is quite adequate for many tasks.Yes, the interactions between electrons and crystall lattice are not exactly like a pinball machine, and the processes giong on in the battery are not as simple as constant external force acting on charge carriers but the net effect is pretty much the same. So you won't get the voltage of a lithium cell or the resistivity of copper accurately. No big deal. You still get all the curcuit laws out of it.

Talking about potential and charges, do you agree that voltage between points A and B (in the absence of magnetic field) is $V_{AB} = \frac{1}{4πε_0} \int_v \rho(\textbf{r}) \left( \frac{1}{|\textbf{r}-B|} - \frac{1}{|\textbf{r}-A|} \right) d\textbf{r}$ ? This just goes to show that voltage is completely determined by charge distribution alone, be it an isolated charge in electrostatics or resistor in circuit or anything else. Thankfully, we do not need to compute this integral, but saying that the charge has nothing to do with voltage is just wrong, there is a clear causal link.

And finally, where, in your opinion, did the 1KV/m electic field came from in my example of 1mm 1Ω resistor?

Regards, DK

---- edit ----

Rainer Müller, "A semiquantitative treatment of surface charges in DC circuits", Am. J. Phys. 80, 782(2012)

ABSTRACT
Surface charges play a major role in DC circuits because they help generate the electric field and potential distributions necessary to move the charges around the circuit.

 

[member 392791]

The thing that amazes me from reading this thread is how people who have studied physics for 20+ years can still argue about the most fundamental of concepts. That is a true testament to the rigor of physics :)

 

[sophiecentaur]

The thing that amazes me from reading this thread is how people who have studied physics for 20+ years can still argue about the most fundamental of concepts. That is a true testament to the rigor of physics :)

Good point. Mind you, they will argue just as much about fishing, football, politics . . . . .

 

[member 392791]

Only the difference is that those things are more subjective, whereas physics has one true answer

 

[sophiecentaur]

Here we go, 2nd hit on google, with pictures:
http://www.astrophysik.uni-kiel.de/~hhaertel/PUB/voltage_IRL.pdf
Sorry, I don't think I can find an article in peer-reviewed journal because frankly, it has all been sorted out a century ago, now it's kid's textbook level stuff.

And again, I'm not suggesting any kind of "alternative approach to electricity", oh no, certainly not. I'm just saying that there is a qualitative picture of the process, based on a simple model, to complement quantitative methods using conservation laws.

It's a bit like colliding billiard balls. Ordinarily one solves the problem by applying energy and momentum conservation. The forces are not easily quantifiable and do not appear anywhere in the equations used to solve the problem. Nevertheless, if the question was "why did the ball change direction" the answer would be "because of the force acting on it" in line with the Newton's 1st law.

And btw, this picture, while simplified, is quite adequate for many tasks.Yes, the interactions between electrons and crystall lattice are not exactly like a pinball machine, and the processes giong on in the battery are not as simple as constant external force acting on charge carriers but the net effect is pretty much the same. So you won't get the voltage of a lithium cell or the resistivity of copper accurately. No big deal. You still get all the curcuit laws out of it.

Talking about potential and charges, do you agree that voltage between points A and B (in the absence of magnetic field) is $V_{AB} = \frac{1}{4πε_0} \int_v \rho(\textbf{r}) \left( \frac{1}{|\textbf{r}-B|} - \frac{1}{|\textbf{r}-A|} \right) d\textbf{r}$ ? This just goes to show that voltage is completely determined by charge distribution alone, be it an isolated charge in electrostatics or resistor in circuit or anything else. Thankfully, we do not need to compute this integral, but saying that the charge has nothing to do with voltage is just wrong, there is a clear causal link.

And finally, where, in your opinion, did the 1KV/m electic field came from in my example of 1mm 1Ω resistor?

Regards, DK

---- edit ----

Rainer Müller, "A semiquantitative treatment of surface charges in DC circuits", Am. J. Phys. 80, 782(2012)

I think most of you guys have missed my main point. There is no doubt that charges are involved in electric current through 'matter'. There is also no doubt (is there?) that Maxwells equations hold in free space - so there is an effective current without the need for electrons to me moving about. That is all well established stuff. But that was not my point.
My argument was against the statement that it is Charge imbalance that causes a current to flow. Now, if you want to say that it is surplus charges at each end of the resistor that cause the current, you surely (?) need to be able to describe some sort of quantitative relationship between those charges. There is not one. The surplus charge imbalance are a result of the Potential Difference and its value can be calculated in terms of the Capacitance involved. The Current in the Resistor is governed by the PD across the resistor and the energy dissipated per Coulomb is equal to that PD.
The PD is the common factor (independent variable) between two quantities (dependent variables) that are not directly related to each other at all. (Any combination of Q and I is possible) If there's no direct relationship then I cannot see how anyone can claim a causal relationship i.e."Current depends on the Charge difference" (or words to that effect).

This stuff is ancient knowledge and not in modern papers. Old research papers will not be available on line. However, there are a million textbooks out there and I challenge anyone to find that inaccurate idea in any book that's aimed at serious students of Physics. For those who do not 'do' theory, perhaps they could look at Spice (or some other simulation) and try to simulate some circuit in which such a relationship could be shown (when the appropriate variables are controlled).

What was it that set up the original Charge distribution in the situation described? Surely it was energy put into the system and established the Potential Energy situation. If there had been no energy input (and that applies to a circuit and to an 'electrostatic' setup) there would be no charge imbalance. This is not just a 'chicken or egg' argument. There is a inherent order of cause and effect.

That would come from the PD and the length of the resistor . But that assumes a straight resistor. If it were horseshoe shaped then the Field (volts per metre) would be irrelevant. The PD alone would be what determines the current. This is a case where Field is just not as useful as Potential.

 

[sophiecentaur]

Only the difference is that those things are more subjective, whereas physics has one true answer

Not sure about that. We may be making progress in the right direction but it is only a matter of Faith that we can ever aspire to the Truth. The answer that works near enough so that we can't detect significant error, is all we can demand.